Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x+6y &= 5 \\ 7x-4y &= -6\end{align*}$
Solution: Begin by moving the $y$ -term in the second equation to the right side of the equation. $7x = 4y-6$ Divide both sides by $7$ to isolate $x$ $x = {\dfrac{4}{7}y - \dfrac{6}{7}}$ Substitute this expression for $x$ in the first equation. $-8({\dfrac{4}{7}y - \dfrac{6}{7}}) + 6y = 5$ $-\dfrac{32}{7}y + \dfrac{48}{7} + 6y = 5$ Simplify by combining terms, then solve for $y$ $\dfrac{10}{7}y + \dfrac{48}{7} = 5$ $\dfrac{10}{7}y = -\dfrac{13}{7}$ $y = -\dfrac{13}{10}$ Substitute $-\dfrac{13}{10}$ for $y$ in the top equation. $-8x+6( -\dfrac{13}{10}) = 5$ $-8x-\dfrac{39}{5} = 5$ $-8x = \dfrac{64}{5}$ $x = -\dfrac{8}{5}$ The solution is $\enspace x = -\dfrac{8}{5}, \enspace y = -\dfrac{13}{10}$.